3.155 \(\int \frac{x^4 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=69 \[ \frac{\sqrt{b x^2+c x^4} (2 b B-A c)}{b c^2 x}-\frac{x^3 (b B-A c)}{b c \sqrt{b x^2+c x^4}} \]

[Out]

-(((b*B - A*c)*x^3)/(b*c*Sqrt[b*x^2 + c*x^4])) + ((2*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(b*c^2*x)

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Rubi [A]  time = 0.149797, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2037, 1588} \[ \frac{\sqrt{b x^2+c x^4} (2 b B-A c)}{b c^2 x}-\frac{x^3 (b B-A c)}{b c \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(((b*B - A*c)*x^3)/(b*c*Sqrt[b*x^2 + c*x^4])) + ((2*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(b*c^2*x)

Rule 2037

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> -Si
mp[(e^(j - 1)*(b*c - a*d)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*b*n*(p + 1)), x] - Dist[(e^j*(a*
d*(m + j*p + 1) - b*c*(m + n + p*(j + n) + 1)))/(a*b*n*(p + 1)), Int[(e*x)^(m - j)*(a*x^j + b*x^(j + n))^(p +
1), x], x] /; FreeQ[{a, b, c, d, e, j, m, n}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && Lt
Q[p, -1] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{x^4 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac{(b B-A c) x^3}{b c \sqrt{b x^2+c x^4}}+\frac{(2 b B-A c) \int \frac{x^2}{\sqrt{b x^2+c x^4}} \, dx}{b c}\\ &=-\frac{(b B-A c) x^3}{b c \sqrt{b x^2+c x^4}}+\frac{(2 b B-A c) \sqrt{b x^2+c x^4}}{b c^2 x}\\ \end{align*}

Mathematica [A]  time = 0.0229948, size = 35, normalized size = 0.51 \[ \frac{x \left (-A c+2 b B+B c x^2\right )}{c^2 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(2*b*B - A*c + B*c*x^2))/(c^2*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.003, size = 44, normalized size = 0.6 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -Bc{x}^{2}+Ac-2\,Bb \right ){x}^{3}}{{c}^{2}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)

[Out]

-(c*x^2+b)*(-B*c*x^2+A*c-2*B*b)*x^3/c^2/(c*x^4+b*x^2)^(3/2)

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Maxima [A]  time = 1.21333, size = 53, normalized size = 0.77 \begin{align*} \frac{{\left (c x^{2} + 2 \, b\right )} B}{\sqrt{c x^{2} + b} c^{2}} - \frac{A}{\sqrt{c x^{2} + b} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

(c*x^2 + 2*b)*B/(sqrt(c*x^2 + b)*c^2) - A/(sqrt(c*x^2 + b)*c)

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Fricas [A]  time = 1.27892, size = 88, normalized size = 1.28 \begin{align*} \frac{\sqrt{c x^{4} + b x^{2}}{\left (B c x^{2} + 2 \, B b - A c\right )}}{c^{3} x^{3} + b c^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

sqrt(c*x^4 + b*x^2)*(B*c*x^2 + 2*B*b - A*c)/(c^3*x^3 + b*c^2*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \left (A + B x^{2}\right )}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**4*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)

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Giac [A]  time = 1.33152, size = 81, normalized size = 1.17 \begin{align*} -\frac{2 \, B \sqrt{b}}{{\left ({\left (\sqrt{c + \frac{b}{x^{2}}} - \frac{\sqrt{b}}{x}\right )}^{2} - c\right )} c} + \frac{B b - A c}{\sqrt{c + \frac{b}{x^{2}}} c^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

-2*B*sqrt(b)/(((sqrt(c + b/x^2) - sqrt(b)/x)^2 - c)*c) + (B*b - A*c)/(sqrt(c + b/x^2)*c^2*x)